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Trigonometric Ratio and Identites question

2004 · Shift 0 · Q75

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JEE Main
Mathematics
Trigonometric Ratio and Identites
2004 · Shift 0 · Q75

Trigonometric Ratio and Identites question

2004 · Shift 0 · Q75

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1

If

u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }

then the difference between the maximum and minimum values of

{u^2}

is given by :

A
${\left( {a - b} \right)^2}$
B
$2\sqrt {{a^2} + {b^2}}$
C
${\left( {a + b} \right)^2}$
D
$2\left( {{a^2} + {b^2}} \right)$

View written solutionFree

Correct answer: A

Let $x=\sqrt{a^2\cos^2\theta+b^2\sin^2\theta},\qquad y=\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}.$ Then $u=x+y.$ We need the maximum and minimum values of $u^2$ .
First compute $u^2$ : $u^2=(x+y)^2=x^2+y^2+2xy.$ Now, $x^2+y^2=(a^2\cos^2\theta+b^2\sin^2\theta)+(a^2\sin^2\theta+b^2\cos^2\theta)=a^2+b^2.$ So, $u^2=a^2+b^2+2xy.$
Now evaluate $xy$ : $xy=\sqrt{(a^2\cos^2\theta+b^2\sin^2\theta)(a^2\sin^2\theta+b^2\cos^2\theta)}.$ Let

\qquad s=\sin^2\theta, \qquad c+s=1.$$ Then $$(a^2c+b^2s)(a^2s+b^2c)=a^4cs+b^4cs+a^2b^2(c^2+s^2).$$ Using $$c^2+s^2=(c+s)^2-2cs=1-2cs,$$ we get $$(a^2c+b^2s)(a^2s+b^2c)=a^2b^2+(a^2-b^2)^2cs.$$ Hence $$xy=\sqrt{a^2b^2+(a^2-b^2)^2\sin^2\theta\cos^2\theta}.$$ Therefore, $$u^2=a^2+b^2+2\sqrt{a^2b^2+(a^2-b^2)^2\sin^2\theta\cos^2\theta}.$$ 4. Since $$\sin^2\theta\cos^2\theta\in\left[0,\frac14\right],$$ the expression inside the square root varies between: - minimum when $\sin^2\theta\cos^2\theta=0$ - maximum when $\sin^2\theta\cos^2\theta=\frac14$ So: ### Minimum of $u^2$ When $\sin^2\theta\cos^2\theta=0$, $$u^2_{\min}=a^2+b^2+2|ab|.$$ Assuming $a,b>0$ (standard convention), $$u^2_{\min}=a^2+b^2+2ab=(a+b)^2.$$ ### Maximum of $u^2$ When $\sin^2\theta\cos^2\theta=\frac14$, $$u^2_{\max}=a^2+b^2+2\sqrt{a^2b^2+\frac{(a^2-b^2)^2}{4}}.$$ Now simplify the square root: $$a^2b^2+\frac{(a^2-b^2)^2}{4} =\frac{4a^2b^2+a^4-2a^2b^2+b^4}{4} =\frac{(a^2+b^2)^2}{4}.$$ Thus, $$u^2_{\max}=a^2+b^2+2\cdot \frac{a^2+b^2}{2}=2(a^2+b^2).$$ 5. Therefore the difference between maximum and minimum values of $u^2$ is $$u^2_{\max}-u^2_{\min}=2(a^2+b^2)-(a+b)^2=a^2+b^2-2ab=(a-b)^2.$$ 6. Hence the correct option is $$\boxed{A:\ (a-b)^2}.$$

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